Solve a with the 1500 m resistor gives with the

Info icon This preview shows pages 5–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Solve: (a) With the 1500 M resistor, gives With the resistor, gives Subtracting the second equation from the first gives and (b) Now R is 7.00 . Or, Reflect: The smaller R is, the larger the current and the smaller the terminal voltage. With the 1500 M resistor the current is very small and the terminal voltage differs only very little from the emf of the battery. 19.30. Set Up: When a battery is short-circuited, Solve: (a) (b) (c) r 5 6.00 V 1000 A 5 6.00 3 10 2 3 V 5 6.00 m V r 5 1.50 V 10.0 A 5 0.150 V r 5 E I 5 1.50 V 25.0 A 5 0.0600 V E 2 Ir 5 0. V V ab 5 IR 5 1.91 V. V ab 5 E 2 Ir 5 2.50 V 2 1 0.274 A 21 2.14 V 2 5 1.91 V. I 5 E R 1 r 5 2.50 V 7.00 V 1 2.14 V 5 0.274 A V E 5 1.75 V 1 1 0.350 A 2 r 5 2.50 V r 5 2.14 V . 0.75 V 5 1 0.350 A 2 r 1.75 V 5 E 2 1 0.350 A 2 r . V ab 5 E 2 Ir I 5 V ab R 5 1.75 V 5.00 V 5 0.350 A. 5.00 V 2.50 V 5 E 2 1 1.67 3 10 2 9 A 2 r . V ab 5 E 2 Ir I 5 V ab R 5 2.50 V 1500 3 10 6 V 5 1.67 3 10 2 9 A. V 1 E , r 2 V ab 5 IR . V ab 5 E 2 Ir . 11.3 V. IR 5 1 2.40 A 21 4.70 V 2 5 E 2 Ir 5 12.0 V 2 1 2.40 A 21 0.30 V 2 5 11.3 V. I 5 E r 1 R 5 12.0 V 4.70 V 1 0.30 V 5 2.40 A. E 2 Ir 2 IR 5 0 V 5 IR I 5 0, I 5 0, R 5 V I 5 2.97 V 1.65 A 5 1.80 V . V 2 IR 5 0 r 5 V 2 E I 5 3.08 V 2 2.97 V 1.65 A 5 0.067 V . I 5 1.65 A, V 5 2.97 V V 5 E 2 Ir , E 5 3.08 V. V 5 E 2 Ir . E . 15.0 A . I T 5 15.0 A 1 1 1 0.0020 1 2 2 1 21 100 C° 2 5 12.5 A. I 0 5 T 5 100°C T 5 0°C, I T 5 I 0 1 1 a 1 T 2 T 0 2 . I 0 5 V R 0 R T 5 R 0 3 1 1 a 1 T 2 T 0 24 . I new 5 V p d new 2 4 r L new 5 V p 1 2 d 2 2 4 r 1 2 L 2 5 2 I d new 5 2 d . L new 5 2 L ; Current, Resistance, and Direct-Current Circuits 19-5
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
19.31. Set Up: When current passes through a battery in the direction from the terminal toward the terminal, the terminal voltage of the battery is Solve: (a) gives (b) so Reflect: The total resistance in the circuit is which agrees with the value specified in the problem. 19.32. Set Up: The graph shows when Solve: (a) When (b) When As A qualitative sketch of versus R is sketched in Figure 19.32a. Figure 19.32 (c) If then The graph of I versus R is sketched in Figure 19.32b. when and when is constant, independent of R. The graph of versus R is sketched in Figure 19.32c. 19.33. Set Up: For a resistor, and Solve: (a) (b) 19.34. Set Up: For a resistor, and Solve: (a) (b) 19.35. Set Up: Solve: (a) (b) 19.36. Set Up: Power is energy per unit time. The power delivered by a voltage source is Solve: (a) (b) Energy 5 Pt 5 1 300 W 21 3.0 3 10 2 3 s 2 5 0.90 J P 5 1 25 V 21 12 A 2 5 300 W. P 5 V ab I . Energy 5 1 1.17 W 21 1.5 h 21 3600 s / h 2 5 6320 J P 5 1 9.0 V 21 0.13 A 2 5 1.17 W Energy 5 Pt . P 5 VI . P 5 V 2 R 5 1 120 V 2 2 9.0 3 10 3 V 5 1.6 W V 5 IR 5 1 0.0183 A 21 15 3 10 3 V 2 5 275 V. I 5 Å P R 5 Å 5.0 W 15 3 10 3 V 5 0.0183 A. V 5 IR . P 5 I 2 R 5 V 2 R 5 VI R 5 V I 5 15.0 V 21.8 A 5 0.688 V I 5 P V 5 327 W 15.0 V 5 21.8 A V 5 IR . P 5 VI V ab V ab V ab 5 E ; R S ` . I S 0 R S 0 I S ` I 5 E R . r 5 0 V ab R E (a) R I (b) V ab R E (c) V ab V ab S E . R S ` , V ab 5 0. R 5 0, V ab 5 IR 5 E R R 1 r 5 E 1 1 r / R . r 5 E I 5 5.00 V 2.5 A 5 2.00 V R 5 0, R 5 0. I 5 2.5 A V ab 5 E 2 Ir . I 5 E R 1 r . I 5 24.0 V 6.00 V 5 4.00 A, R 1 r 5 6.00 V . R 5 V ab I 5 21.2 V 4.00 A 5 5.30 V . V ab 2 IR 5 0 r 5 E 2 V ab I 5 24.0 V 2 21.2 V 4.00 A 5 0.700 V . V ab 5 E 2 Ir V ab 5 E 2 Ir . V ab 1 2 19-6 Chapter 19
Image of page 6
19.37.
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern